If a plane flying 60 degrees northeast at 200 km/hr is struck by a wind that blows due west at 50 km/hr, what angle must the plane fly to maintain the sixty degrees northeast bearing.

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To be honest, I really have no idea how to start this.

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- Oct 18th 2009, 02:50 PMCaturdayzWind Resistance Problem
If a plane flying 60 degrees northeast at 200 km/hr is struck by a wind that blows due west at 50 km/hr, what angle must the plane fly to maintain the sixty degrees northeast bearing.

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To be honest, I really have no idea how to start this. - Oct 19th 2009, 05:10 AMaidan
1st: make a sketch

Identify all that is given.

After 1 hour where would the the plane be if there were no problem with the wind?

(Call the starting or origin Point A and call the point where the airplane would be after 1 hour as Point B.)

You need to determine that.

Hint: at a speed of 200km/h with a heading of 60degrees northeast,

determine the km distance NORTH of the starting point

& the km distance EAST of the starting point.

[ N = 200 cos(60); E = 200 sin(60) ]

(The wind (at 50 km/h) will effectively move the airplane 50 km WEST in 1 hour. This is what you need to counteract or correct.)

THEN, from that location where the plane would be after 1 hour (which was point B), move it directly**EAST**50 km (call this point C).

From the starting point (A), the airplane needs to head towards point C, to correct for the wind.

The adjusted bearing is:

(You've already computed the values for N & E above.)

$\displaystyle arc\tan\left(\dfrac{E+50}{N}\right)$

. - Dec 14th 2009, 03:47 AMHallsofIvy
Notice that this is NOT a question about "wind resistance" as if, say wind were pushing a sailboat. The airplane is actually

**supported**by the wind just as if you were rolling a toy car over a table while carrying the table to one side. Add the two speeds "vectorially".

One thing you can do is separate into "components". Since cos(60)= $\displaystyle \sqrt{3}/2$, the eastward component of the planes flight, relative to the wind, is $\displaystyle (200)\sqrt{3}/2= 100\sqrt{3}$. Similarly, since sin(60)= 1/2, the northward component is (1/2)(200)= 100.

The wind is blowing directly west so its**east**ward component is -50 while its northward component is 0. Adding those gives a vector whose eastward component is $\displaystyle \100\sqrt{3}- 50$ and whose northward component is $\displaystyle 100$. You can represent that as a right triangle having legs of length $\displaystyle 100\sqrt{3}- 50$ and 100. Thinking of the north component as "opposite side" and the east component as "near side" $\displaystyle tan(\theta)= \frac{100\sqrt{3}- 50}{100}$ where $\displaystyle \theta$ is the angle the direction makes "north of east" or simply "northeast".